26

我正在用复杂的分层信息抓取一些数据，需要将结果导出到 json。

我将项目定义为

class FamilyItem():
    name = Field()
    sons = Field()

class SonsItem():
    name = Field()
    grandsons = Field()

class GrandsonsItem():
    name = Field()
    age = Field()
    weight = Field()
    sex = Field()

当蜘蛛运行完成时，我会得到一个打印的项目输出，比如

{'name': 'Jenny',
   'sons': [
            {'name': u'S1',
             'grandsons': [
                   {'name': u'GS1',
                    'age': 18,
                    'weight': 50
                   },
                   {
                    'name':u'GS2',
                    'age': 19,
                    'weight':51}]
                   }]
}

但是当我运行时scrapy crawl myscaper -o a.json，它总是说结果“不是 JSON 可序列化的”。然后我将项目输出复制并粘贴到 ipython 控制台并使用 json.dumps()，它工作正常。那么问题出在哪里？这让我发疯了……

4

2 回答 2

34

保存嵌套项时，请确保将它们包装在对 dict() 的调用中，例如：

gs1 = GrandsonsItem()
gs1['name'] = 'GS1'
gs1['age'] = 18
gs1['weight'] = 50

gs2 = GrandsonsItem()
gs2['name'] = 'GS2'
gs2['age'] = 19
gs2['weight'] = 51

s1 = SonsItem()
s1['name'] = 'S1'
s1['grandsons'] = [dict(gs1), dict(gs2)]

jenny = FamilyItem()
jenny['name'] = 'Jenny'
jenny['sons'] = [dict(s1)]
于 2012-12-19T02:27:44.807 回答
2

不确定是否有办法在scrapy中使用类做嵌套项目，但数组工作正常。你可以这样做：

grandson = Grandson(name = 'Grandson', age = 2)

son = Son(name = 'Son', grandsons = [grandson])

item = Item(name = 'Name', son = [son])
于 2012-06-26T13:13:23.430 回答